Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; i < (int)(n); i++)
using namespace std;
using ll = long long;

template <typename T>
struct Dijkstra {
  struct Edge {
    int to;
    T cost;
  };
  vector<int> prev;
  vector<vector<Edge>> g;
  Dijkstra(int n) : prev(n, -1), g(n) {}

  void addEdge(int u, int v, T w) {
    g[u].push_back({v, w});
    g[v].push_back({u, w});
  }
  vector<T> build(int s) {
    vector<T> dist(g.size(), -1);
    using Node = pair<T, int>;
    priority_queue<Node, vector<Node>, greater<Node>> pq;
    pq.push({dist[s] = 0, s});

    while (!pq.empty()) {
      auto d = pq.top().first;
      auto u = pq.top().second;
      pq.pop();
      if (dist[u] < d) continue;
      for (auto&& v : g[u]) {
        if (dist[v.to] < 0 || dist[v.to] > dist[u] + v.cost) {
          dist[v.to] = dist[u] + v.cost;
          prev[v.to] = u;
          pq.push({dist[v.to], v.to});
        }
      }
    }
    return dist;
  }
  vector<int> getPath(int t) {
    vector<int> path;
    for (; t != -1; t = prev[t]) path.push_back(t);
    reverse(begin(path), end(path));
    return path;
  }
};

int main() {
  int n, m, s, t;
  cin >> n >> m >> s >> t, s--, t--;
  Dijkstra<int> g(n);
  REP(i, m) {
    int x, y;
    cin >> x >> y, x--, y--;
    g.addEdge(x, y, 1);
  }
  auto ds = g.build(s);
  auto dt = g.build(t);
  map<int, int> ms, mt;
  for (auto x : ds) ms[x]++;
  for (auto x : dt) mt[x]++;
  ll ans = 0LL;
  int c = ds[t] - 1;
  REP(i, c) {
    ans += (ll)ms[i] * mt[c - 1 - i];
  }
  cout << ans << endl;
}